Question

In $△ABC,\angle ABC={90}^o$ and BM is the altitude. If AM = 16 MC Prove that $\text{AB}=4\text{BC}$

Answer 1

Given :

$\angle ABC=90°$

$AM=16MC$

To proof : $AB=4BC$

let $MC=x$

$\therefore AM=16x$

In $△AMB$ & $△ABC$

$\angle A=\angle A$ $\left(common\right)$

$\angle AMB=\angle ABC$ $\left(90°\right)$

$△AMB\sim △ABC$ (by $AA$)

$\therefore \frac{AM}{AB}=\frac{BM}{BC}⟹AM\times BC=AB\times BM⟶\left(1\right)$

Now in $△CMB$ & $△CBA$

$\angle C=\angle C$ $\left(common\right)$

$\angle CMB=\angle CBA$ $\left(90°\right)$

$△CMB\sim △CBA$ (by $AA$)

$\therefore \frac{CM}{CB}=\frac{MB}{AB}⟹AB\times CM=MB\times BC⟶\left(2\right)$

Divide $\left(1\right)$ by $\left(2\right)$

$\frac{AM\times BC}{AB\times CM}=\frac{AB\times BM}{MB\times BC}$

$[\because AM=16CM]$ $\frac{16\times CM\times BC}{AB\times CM}=\frac{AB}{BC}⟹16{BC}^2={AB}^2$

$AB=4BC$